Force on a Current-Carrying Conductor in a Magnetic Field

Consider a straight conductor of length l and cross-sectional area A, carrying current i. It is placed in a field of magnetic induction B at right angles to the length of the conductor and directed into the plane of the paper.

The electric current, I, in a conductor is due to free electrons moving in a direction opposite to the direction of the electric current. Let n be the number of free electrons per unit volume of the conductor and vd the drift velocity of the electrons.



The magnitude of the force on each electron is
        f = evd B.        (. : vd  _|_ B)

The number of electrons in the length l of the conductor is
        N = n l A.

Therefore, total force on all the free electrons, i.e., on the length l of the conductor is
    F = fN = (e vd B) N = (e vd B) (nlA) = (n Ae vd ) (Bl).

But n A e vd = i, the current n the wire
. : F = I Bl.
If the conductor makes an angle θ with the magnetic field induction B, then the force is
    F = iBl sin θ.
or     Vectorially, F = il x B.

Here, I is a vetor pointing along the straight conductor in the direction of the current.
Direction of F is perpendicular to both, I and B and is given by Fleming’s left hand rule.



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